How do you differentiate f(x)=e^((6x-2)^2 f(x)=e(6x2)2 using the chain rule.?

1 Answer
Apr 6, 2016

Answer:
h'(x)= 12(6x-2)*e^((6x-2)^2)=(72x-24)*e^((6x-2)^2)

Explanation:

Given : f(x)=e^((6x-2)^2

Required: The derivative of f(x)

Definition and principles: The chain rule
Let h(x)=(f@g)(x) then the derivative of h(x) is

h'(x)= f'(g(x))g'(x)

Solution Strategy: let f(x)=(6x-2)^2 and g(x)=e^(f(x))
A) Evaluate f'(g(x)) and g'(x) B) Put it all together - f'(g(x))*g'(x) #

A) I am going to write the chain rule as:
let g(x) = (6x-2)^2, then g'(x)= 12(6x-2)
now f(g(x))= e^(g(x)) and f'(g(x)) = e^g(x)=color(red)(e^((6x-2)^2)
B) f'(g(x))*g'(x) = color(red)(e^((6x-2)^2)*color(blue)(12(6x-2))

Answer:
h'(x)= 12(6x-2)*e^((6x-2)^2)=(72x-24)*e^((6x-2)^2)

Remark: for the chain rule you can use the notation given below which I find a little simpler to follow and is the notation most frequently used especially in physics community.

(dh(x))/(dx)=(df(u))/(du)* (du)/(dx) and you let
u=(6x-2)^2 and f(u)=e^u then
(df(u))/(du)=e^u and (du)/(dx)=12(6x-2)
and (dh(x))/(dx)=[e^u*12(6x-2)]_(u=(6x-2)^2)=(72x-24)*e^((6x-2)^2)